6 Most VSAQ’s of Mean Value Theorems Chapter in Inter 1st Year Maths-1B (TS/AP)

2 Marks

VSAQ-1 : State Rolle’s Mean value theorem

If a function f(x) is

(i) continuous on [a,b]

(ii) derivable on (a,b)

(iii) f(a) = f(b) then there exists c ∈ (a,b) such that f'(c) = 0

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VSAQ-2 : State Lagrange’s Mean value theorem

If a function f(x) is

(i) Continuous on [a,b]

(ii) deriivable on (a,b) then there exists c ∈ (a,b) such that

$$[
f'(c) = \frac{f(b) – f(a)}{b – a}
]$$

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VSAQ-3 : Verify Rolle’s theorem for the function y = f(x) = x² + 4 on [-3,3]

$$f(x) = x^2 + 4$$

$$f'(x) = 2x$$

$$\text{(i) continuous on } [-3,3]$$

$$\text{(ii) differentiable in } (-3,3)$$

$$f(-3) = (-3)^2 + 4 = 9 + 4 = 13$$

$$f(3) = 3^2 + 4 = 9 + 4 = 13$$

$$f(-3) = f(3)$$

$$\text{So, from Rolle’s theorem } f'(c) = 0$$

$$2c = 0$$

$$c = 0$$

$$c = 0 \in (-3,3)$$

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VSAQ-4 : Verify Rolle’s theorem for the function x² – 5x + 6 on |-3,8|

$$f(x) = x^2 – 5x + 6$$

$$f'(x) = 2x – 5$$

$$\text{(i) continuous on } [-3,8]$$

$$\text{(ii) differentiable in } (-3,8)$$

$$f(-3) = (-3)^2 – 5(-3) + 6 = 9 + 15 + 6 = 30$$

$$f(8) = 8^2 – 5 \times 8 + 6 = 64 – 40 + 6 = 30$$

$$f(-3) = f(8)$$

$$\text{So, from Rolle’s theorem } f'(c) = 0$$

$$2c – 5 = 0$$

$$2c = 5$$

$$c = \frac{5}{2} = 2.5$$

$$c = 2.5 \in (-3,8)$$

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VSAQ-5 : Verify the conditions of Lagrange’s mean value theorem for the function x² – 1 on |2,3|

$$f(x) = x^2 – 1$$

$$f'(x) = 2x$$

$$\text{(i) continuous on } [2,3]$$

$$\text{(ii) differentiable in } (2,3)$$

$$\text{So, from Lagrange’s theorem }$$

$$f'(c) = \frac{f(3) – f(2)}{3 – 2}$$

$$f'(c) = \frac{(3^2 – 1) – (2^2 – 1)}{3 – 2} = \frac{(9 – 1) – (4 – 1)}{1} = \frac{8 – 3}{1} = 5$$

$$2c = 5$$

$$c = \frac{5}{2} = 2.5$$

$$c = 2.5 \in (2,3)$$

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VSAQ-6 : Verify Lagrange’s mean value theorem for the function f(x) = x² on [2,4]

$$f(x) = x^2$$

$$f'(x) = 2x$$

$$\text{(i) continuous on } [2,4]$$

$$\text{(ii) differentiable in } (2,4)$$

$$\text{So, from Lagrange’s theorem,}$$

$$f'(c) = \frac{f(4) – f(2)}{4 – 2}$$

$$f'(c) = \frac{4^2 – 2^2}{4 – 2} = \frac{16 – 4}{2} = \frac{12}{2} = 6$$

$$2c = 6$$

$$c = 3$$

$$c = 3 \in (2,4)$$